Capacitance and the parallel plate capacitorAQA A-Level Physics: Revision notes

A capacitor stores charge. The capacitance C is the charge stored per unit potential difference across it:

$C = \frac{Q}{V}$

The unit is the farad (F), equal to one coulomb per volt (C V⁻¹). Q is the magnitude of the charge on one plate; the net charge on the capacitor is zero. Typical values are microfarads (µF, 10⁻⁶ F) to millifarads.

Worked example. A 220 µF capacitor charged to 9.0 V stores $Q = CV = 220 \times 10^{-6} \times 9.0 = 2.0 \times 10^{-3}$ C.

For two parallel plates of area A separated by a distance d, with empty space (or air) between them:

$C = \frac{A\varepsilon_0}{d}$

So C is proportional to A and inversely proportional to d. Doubling the area doubles C; doubling the separation halves it.

Worked example. A = 0.010 m², d = 2.0 mm: $C = \dfrac{8.85 \times 10^{-12} \times 0.010}{2.0 \times 10^{-3}} = 4.4 \times 10^{-11}$ F (44 pF).

A dielectric is an insulating material placed between the plates. It increases the capacitance. With a dielectric filling the gap:

$C = \frac{A\varepsilon_0\varepsilon_r}{d}$

The relative permittivity εᵣ (also called the dielectric constant) is the factor by which the capacitance is multiplied when the dielectric replaces a vacuum or air:

$\varepsilon_r = \frac{C_{\text{with dielectric}}}{C_{\text{without}}}$

εᵣ has no unit. The product ε = ε₀εᵣ is the permittivity of the dielectric.

Worked example. The capacitor above is filled with a material of εᵣ = 5.0. New C = 5.0 × 44 pF = 2.2 × 10⁻¹⁰ F.

Some dielectrics contain polar molecules: molecules with a positive end and a negative end, such as water. Without a field they point in random directions.

When the capacitor is charged:

1. The molecules rotate so that their positive ends point towards the negative plate and their negative ends towards the positive plate.
2. The aligned molecules produce a field that opposes the field due to the plates. Equivalently, charge of the opposite sign builds up on each surface of the dielectric.
3. The net field, and so the p.d., is reduced for the same charge on the plates.
4. Since $C = Q/V$, a smaller V for the same Q means a larger capacitance.

A 1.5 nF capacitor is charged to 80 V and disconnected. A dielectric is inserted and the capacitance becomes 6.0 nF.

• $\varepsilon_r = 6.0 / 1.5 = 4.0$
• Q is fixed: $Q = 1.5 \times 10^{-9} \times 80 = 1.2 \times 10^{-7}$ C
• New p.d.: $V = Q/C = 1.2 \times 10^{-7} / 6.0 \times 10^{-9} = 20$ V

The p.d. falls by the same factor (4.0) as the capacitance rises.